Consider two conductinbg spheres of radill `R_(1) and R_(2)` with `R_(1)gt R_(2).` If the two are at the same potential, and the larger sphere has more charge than the smaller sphere, then

Options

A. the charge density of smaller sphere is less then that of larger sphere,
B. the charge density of smaller sphere is more than that of larger sphere.
C. both spheres may have same charge density.
D. none of these

Explanation

To solve the problem, we need to analyze the situation involving two conducting spheres with different radii and charges. Let's break down the solution step by step.

### Step 1: Understand the Given Information
We have two conducting spheres:
- Sphere 1 with radius \( R_1 \) (larger sphere)
- Sphere 2 with radius \( R_2 \) (smaller sphere)

It is given that \( R_1 > R_2 \) and that both spheres are at the same electric potential. Additionally, Sphere 1 has more charge than Sphere 2.

### Step 2: Write the Expression for Electric Potential
The electric potential \( V \) of a charged conducting sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the sphere.

For the two spheres, we can write:
\[ V_1 = \frac{kQ_1}{R_1} \quad \text{(for Sphere 1)} \]
\[ V_2 = \frac{kQ_2}{R_2} \quad \text{(for Sphere 2)} \]

### Step 3: Set the Potentials Equal
Since both spheres are at the same potential, we can set \( V_1 = V_2 \):
\[ \frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \]

### Step 4: Simplify the Equation
We can cancel \( k \) from both sides:
\[ \frac{Q_1}{R_1} = \frac{Q_2}{R_2} \]

### Step 5: Rearrange to Find the Relationship Between Charges
Rearranging gives us:
\[ Q_1 = Q_2 \cdot \frac{R_1}{R_2} \]

### Step 6: Analyze the Charges
Since it is given that \( Q_1 > Q_2 \) (the larger sphere has more charge), we can conclude that:
\[ \frac{R_1}{R_2} > 1 \quad \text{(because \( R_1 > R_2 \))} \]
This means that \( Q_1 > Q_2 \) is consistent with the relationship we derived.

### Step 7: Consider Charge Densities
The surface charge density \( \sigma \) for a sphere is given by:
\[ \sigma = \frac{Q}{4\pi R^2} \]
Thus, for the two spheres:
\[ \sigma_1 = \frac{Q_1}{4\pi R_1^2} \quad \text{(for Sphere 1)} \]
\[ \sigma_2 = \frac{Q_2}{4\pi R_2^2} \quad \text{(for Sphere 2)} \]

### Step 8: Relate Charge Densities
From the relationship \( Q_1 = Q_2 \cdot \frac{R_1}{R_2} \), we can substitute \( Q_1 \) into the expression for \( \sigma_1 \):
\[ \sigma_1 = \frac{Q_2 \cdot \frac{R_1}{R_2}}{4\pi R_1^2} \]
Now, we can express \( \sigma_1 \) in terms of \( \sigma_2 \):
\[ \sigma_1 = \frac{Q_2}{4\pi R_2^2} \cdot \frac{R_1}{R_1} \cdot \frac{R_1}{R_2} = \sigma_2 \cdot \frac{R_1}{R_2} \cdot \frac{1}{R_1} \]
This simplifies to:
\[ \sigma_1 = \sigma_2 \cdot \frac{R_1}{R_2} \]

### Step 9: Compare Charge Densities
Since \( R_1 > R_2 \), it follows that:
\[ \sigma_1 < \sigma_2 \]
Thus, the charge density on the smaller sphere is greater than that on the larger sphere.

### Conclusion
The final conclusion is that the charge density on the smaller sphere is greater than that on the larger sphere.